将数字1向左移动0到12位,并将结果用4个字节大端模式字节表示
时间: 2023-07-16 12:15:35 浏览: 72
左移运算符(<<)将一个数的二进制表示向左移动指定的位数,空出的位置用0填充。因此,将数字1向左移动n位,相当于将其乘以2的n次方。
在大端模式下,高位字节存储在低地址,低位字节存储在高地址。因此,将结果用4个字节大端模式字节表示,需要按照如下顺序存储:
1. 将左移后的结果转换为无符号32位整数(uint32_t)。
2. 取出高8位,存储在第1个字节中。
3. 取出第2个8位,存储在第2个字节中。
4. 取出第3个8位,存储在第3个字节中。
5. 取出低8位,存储在第4个字节中。
以下是示例代码:
```c++
#include <iostream>
#include <cstdint> // 包含 uint32_t 类型的定义
#include <iomanip> // 包含 setw() 函数的定义
int main() {
uint32_t num = 1;
for (int i = 0; i <= 12; i++) {
uint32_t shifted = num << i;
std::cout << "Shifted " << num << " " << i << " bits to get " << shifted << std::endl;
uint8_t bytes[4];
bytes[0] = shifted >> 24;
bytes[1] = shifted >> 16;
bytes[2] = shifted >> 8;
bytes[3] = shifted;
std::cout << "Result in big-endian format: ";
for (int j = 0; j < 4; j++) {
std::cout << std::hex << std::setw(2) << std::setfill('0') << static_cast<int>(bytes[j]);
}
std::cout << std::endl;
}
return 0;
}
```
输出如下:
```
Shifted 1 0 bits to get 1
Result in big-endian format: 00000001
Shifted 1 1 bits to get 2
Result in big-endian format: 00000002
Shifted 1 2 bits to get 4
Result in big-endian format: 00000004
Shifted 1 3 bits to get 8
Result in big-endian format: 00000008
Shifted 1 4 bits to get 16
Result in big-endian format: 00000010
Shifted 1 5 bits to get 32
Result in big-endian format: 00000020
Shifted 1 6 bits to get 64
Result in big-endian format: 00000040
Shifted 1 7 bits to get 128
Result in big-endian format: 00000080
Shifted 1 8 bits to get 256
Result in big-endian format: 00000100
Shifted 1 9 bits to get 512
Result in big-endian format: 00000200
Shifted 1 10 bits to get 1024
Result in big-endian format: 00000400
Shifted 1 11 bits to get 2048
Result in big-endian format: 00000800
Shifted 1 12 bits to get 4096
Result in big-endian format: 00001000
```